Cho tam giác ABC có A(-1,1);B(5,-3),đỉnh C nằm trên trục oy và trọng tâm G nằm trên trục ox.Tìm toạ độ điểm C
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NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2023
Đề chính xác là \(\left|\overrightarrow{EB}-3\overrightarrow{EC}\right|\) đạt min đúng ko?
\(\overrightarrow{AB}=\left(2;-4\right)=2\left(1;-2\right)\) nên đường thẳng AB nhận \(\left(2;1\right)\) là 1 vtpt
Phương trình AB:
\(2\left(x+2\right)+1\left(y-1\right)=0\Leftrightarrow2x+y+3=0\)
Do E thuộc AB, đặt \(E\left(a;b\right)\Rightarrow2a+b+3=0\Rightarrow b=-2a-3\)
\(\Rightarrow E\left(a;-2a-3\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{EB}=\left(-a;2a\right)\\\overrightarrow{EC}=\left(1-a;2a+4\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{EB}-3\overrightarrow{EC}=\left(2a-3;-4a-12\right)\)
\(\Rightarrow\left|\overrightarrow{EB}-3\overrightarrow{EC}\right|=\sqrt{\left(2a-3\right)^2+\left(-4a-12\right)^2}=\sqrt{20a^2+84a+153}\)
\(=\sqrt{20\left(a+\dfrac{21}{10}\right)^2+\dfrac{324}{5}}\ge\sqrt{\dfrac{324}{5}}\)
Dấu = xảy ra khi \(a+\dfrac{21}{10}=0\Rightarrow a=-\dfrac{21}{10}\)
\(\Rightarrow E\left(-\dfrac{21}{10};\dfrac{6}{5}\right)\)
30 tháng 12 2021
\(\overrightarrow{CA}=\left(1-x_C;-2\right)\)
\(\overrightarrow{CB}=\left(-2-x_C;2\right)\)
\(\Leftrightarrow\left(x_C-1\right)\left(x_C+2\right)-4=0\)
\(\Leftrightarrow x_C^2+x_C-6=0\)
hay \(x_C=-3\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2022
Áp dụng công thức trọng tâm:
\(\left\{{}\begin{matrix}x_C=3x_G-x_A-x_B=-3\\y_C=3y_G-y_A-y_B=-4\end{matrix}\right.\) \(\Rightarrow C\left(-3;-4\right)\)
\(\Rightarrow\overrightarrow{CA}=\left(4;5\right)\) ; \(\overrightarrow{AB}=\left(1;2\right)\)
Đường cao d đi qua B vuông góc AC nên nhận \(\overrightarrow{CA}=\left(4;5\right)\) là 1 vtpt
Phương trình d:
\(4\left(x-2\right)+5\left(y-3\right)=0\Leftrightarrow4x+5y-23=0\)
Đường cao d1 đi qua C vuông góc AB nên nhận (1;2) là 1 vtpt
Phương trình d1:
\(1\left(x+3\right)+2\left(y+4\right)=0\Leftrightarrow x+2y+11=0\)
H là giao điểm d và d1 nên tọa độ thỏa mãn:
\(\left\{{}\begin{matrix}4x+5y-23=0\\x+2y+11=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{101}{3}\\y=-\dfrac{67}{3}\end{matrix}\right.\) \(\Rightarrow H\left(\dfrac{101}{3};-\dfrac{67}{3}\right)\)
15 tháng 6 2022
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Do C thuộc trục Oy nên tọa độ có dạng \(C\left(0;c\right)\)
Áp dụng công thức trọng tâm:
\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{4}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{c-2}{3}\end{matrix}\right.\)
Do G thuộc Ox \(\Rightarrow y_G=0\Rightarrow\dfrac{c-2}{3}=0\Rightarrow c=2\)
\(\Rightarrow C\left(0;2\right)\)